Does there exist a polynomial $p(x) \in \mathbb C[x]$ such that $p(x) \notin \mathbb R[x]$ and $p(x)p(-x)=p(x^2)$? 
Does there exist a polynomial $p(x) \in \mathbb C[x]$ such that $p(x) \notin \mathbb R[x]$ and $p(x)p(-x)=p(x^2)$ ?

I have noticed that if $a_n$ is the leading coefficient of $p(x)$ then $a_n=(-1)^n $, so the leading coefficient cannot be imaginary; also $p(0)^2=p(0)$ so the constant term also is real. Please help. Thanks in advance.
 A: Here's an example.  Let $\zeta=e^{2\pi i/7}$ and let $p(x)=(\zeta-x)(\zeta^2-x)(\zeta^4-x)$.  Since $\zeta$ is a root of $p$ but $\bar{\zeta}$ is not, $p$ does not have real coefficients.  Now notice that $(\zeta^4)^2=\zeta$, so the squaring map permutes the roots of $p(x)$.  We can use this to compute
\begin{align*}
p(x^2)&=(\zeta-x^2)(\zeta^2-x^2)(\zeta^4-x^2)\\
&=(\zeta^4-x)(\zeta^4+x)(\zeta-x)(\zeta+x)(\zeta^2-x)(\zeta^2+x)\\
&=(\zeta-x)(\zeta^2-x)(\zeta^4-x)(\zeta+x)(\zeta^2+x)(\zeta^4+x)\\
&=p(x)p(-x).
\end{align*}
More generally, if $p(x^2)=p(x)p(-x)$, then the set of roots of $p$ must be closed under squaring, and also for each root of $p$ at least one of its square roots must also be a root of $p$.  Since $p$ can only have finitely many roots, it follows that the only roots $p$ can have are $0$ and roots of unity of odd order.  To arrange that $p$ does not have real coefficients, you then just need to find a collection of roots of unity of odd order which is closed under squaring but not under conjugation.
Explicitly, we can say the following.  Let $\zeta\in\mathbb{C}$ be either $0$ or a root of unity of odd order and let $S(\zeta)=\{\zeta^{2^k}:k\in\mathbb{N}\}$.  Then $S(\zeta)$ is finite and $s\mapsto s^2$ is a bijection from $S(\zeta)$ to itself.  Define $p_\zeta(x)=\prod_{s\in S(\zeta)}(s-x)$.  The same argument as given above for $\zeta=e^{2\pi i/7}$ shows that $p_\zeta(x^2)=p_\zeta(x)p_\zeta(-x)$.  Moreover, $p_\zeta(x)$ has real coefficients iff $\bar{\zeta}\in S(\zeta)$.  (If $\zeta$ is a primitive $n$th root of unity, then $\bar{\zeta}\in S(\zeta)$ iff $-1$ is a power of $2$ mod $n$.)
Conversely, suppose $p(x)$ is any polynomial such that $p(x^2)=p(x)p(-x)$.  Then if $\zeta$ is any root of $p$, as discussed above $\zeta$ must be either $0$ or a root of unity of odd order.  Moreover, since the roots of $p$ are closed under squaring, every element of $S(\zeta)$ is a root of $p$.  It follows that $p$ is divisible by $p_\zeta$, and then $q(x)=p(x)/p_\zeta(x)$ will also satisfy $q(x^2)=q(x)q(-x)$.  By induction on the degree of $p$ (the base case being where $p$ is a constant, which can only be $0$ or $1$), we can thus conclude:

A nonzero polynomial $p(x)\in\mathbb{C}[x]$ satisfies $p(x^2)=p(x)p(-x)$ iff it is a product of polynomials of the form $p_\zeta(x)$ where $\zeta$ is either $0$ or a root of unity of odd order.

